Quantitative Aptitude PYQs

Given:

Original number = x

The resulting number = \(122\frac{1}{2}% of x = \frac{245}{200}x\)

Formula used:

New number = Original number + Added value

Calculation:

\(\frac{245}{200}x\) = x + 18

⇒ \(\frac{245}{200}x - x\) = 18

⇒ \(\frac{45}{200}x\) = 18

⇒ x = \(\frac{18 \times 200}{45}\)

⇒ x = 80

New number = x + 18

⇒ New number = 80 + 18 = 98

∴ The new number is 98.

Given:

Number ranges: 30 to 49, 40 to 60, 21 to 40, and 60 to 80

Formula used:

Prime numbers are numbers greater than 1 that are divisible only by 1 and themselves.

Calculation:

Prime numbers in each range:

30 to 49: 31, 37, 41, 43, 47 (5 primes)

40 to 60: 41, 43, 47, 53, 59 (5 primes)

21 to 40: 23, 29, 31, 37 (4 primes)

60 to 80: 61, 67, 71, 73, 79 (5 primes)

⇒ Range 21 to 40 has the least number of prime numbers.

∴ The correct answer is option (3).

Shortcut Trick

Rate of depreciation (R) = 15% = 3/20.

Value after 4 years = Present Value × (1 − R/100)⁴.

Value = 51,20,000 × (17/20)⁴ = 51,20,000 × 83,521 ÷ 1,60,000.

Value = 32 × 83,521 = ₹26,72,672.

∴ The correct answer is ₹26,72,672.

Alternate Method

Given:

Present Value (P) = ₹51,20,000

Depreciation Rate (r) = 15% per annum

Time Period (n) = 4 years

Formula Used:

Value after n years = P × (1 − r/100)ⁿ

Calculations:

⇒ Future Value = 51,20,000 × (1 − 15/100)⁴

⇒ Future Value = 51,20,000 × (85/100)⁴

⇒ Future Value = 51,20,000 × (17/20) × (17/20) × (17/20) × (17/20)

⇒ Future Value = 51,20,000 × 83,521 ÷ 1,60,000

⇒ Future Value = 32 × 83,521

⇒ Future Value = ₹26,72,672

∴ The correct answer is ₹26,72,672.

Additional Information

Depreciation vs Appreciation

While depreciation uses the formula P(1 − r/100)ⁿ, appreciation (like population growth) uses P(1 + r/100)ⁿ.

Effective Rate of Depreciation

For two years with a rate r%, the net depreciation is given by the successive percentage formula: (r + r − (r × r)/100)%.

Initial Value Calculation

To find the value n years ago given the present value (V), use the formula: Initial Value = V ÷ (1 − r/100)ⁿ.

Given:

Three solid cubes of sides 3 cm, 4 cm, and 5 cm are melted to form a new solid cube.

Formula used:

Volume of a cube = side³

Surface area of a cube = 6 × side²

Calculations:

Volume of the new cube = Volume of the first cube + Volume of the second cube + Volume of the third cube

Volume of the first cube = 3³ = 27 cm³

Volume of the second cube = 4³ = 64 cm³

Volume of the third cube = 5³ = 125 cm³

⇒ Total volume = 27 + 64 + 125

⇒ Total volume = 216 cm³

Let the side of the new cube be x cm.

Then, x³ = 216

⇒ x = 6 cm

Surface area of the new cube = 6 × side²

⇒ Surface area = 6 × 6²

⇒ Surface area = 6 × 36

⇒ Surface area = 216 cm²

∴ The correct answer is option (3).

Given:

An article is marked at a price which gives a profit of 20%. After allowing a certain discount, the profit reduces to 10%.

Formula used:

Discount Percentage = ((Marked Price - Selling Price) / Marked Price) × 100

Calculation:

Let the Cost Price (CP) be 100.

Marked Price (MP) = CP + 20% of CP = 100 + 20 = 120

New Selling Price (SP) with 10% profit = CP + 10% of CP = 100 + 10 = 110

Discount = MP - SP = 120 - 110 = 10

Discount Percentage = (10 / 120) × 100

⇒ Discount Percentage = (1 / 12) × 100

⇒ Discount Percentage = \(8\frac{1}{3}%\)

∴ The correct answer is option 1.

Given:

The ratio of two numbers is 5:6.

Their HCF is 8.

Formula used:

LCM × HCF = Product of the numbers

Calculation:

Let the two numbers be 5x and 6x.

Given HCF = 8, so x = 8.

The numbers are 5×8 = 40 and 6×8 = 48.

Product of the numbers = 40 × 48.

LCM × HCF = 40 × 48.

LCM × 8 = 1920.

⇒ LCM = 1920 ÷ 8.

⇒ LCM = 240.

∴ The correct answer is option (3).

Given:

A sum of ₹18,910 is divided among Arjun, Raima, and Sohan.

Arjun receives 20% less than Raima.

Sohan receives 25% more than Raima's share.

Formula used:

Let Raima's share be R.

Arjun's share = R - 0.20R = 0.80R

Sohan's share = R + 0.25R = 1.25R

Total share = Arjun's share + Raima's share + Sohan's share

Total share = 0.80R + R + 1.25R = 3.05R

Total amount = ₹18,910

3.05R = 18,910

Calculation:

3.05R = 18,910

⇒ R = 18,910 / 3.05

⇒ R = 6,200

Sohan's share = 1.25R

⇒ Sohan's share = 1.25 × 6,200

⇒ Sohan's share = 7,750

∴ The correct answer is option (1).

Given:

The ratio of P's salary to Q's salary is 2 : 3.

The ratio of Q's salary to R's salary is 4 : 5.

Formula used:

To find the ratio of P's salary to R's salary, we need to combine the given ratios.

Calculation:

Let P's salary be 2x, Q's salary be 3x, and R's salary be 5y.

Given Q's salary is equal in both ratios, we equate them:

3x = 4y

⇒ y = 3x / 4

Now, we find the ratio of P's salary to R's salary:

P's salary : R's salary = 2x : 5y

⇒ 2x : 5(3x / 4)

⇒ 2x : (15x / 4)

⇒ 8x : 15x

⇒ 8 : 15

∴ The correct answer is option (4).

Shortcut Trick

For a sum to double itself, the Simple Interest must be equal to the Principal (100%).

Time (T) = 100 ÷ Rate (R)

⇒ T = 100 ÷ 5

⇒ T = 20 years

∴ The correct answer is 20 years.

Alternate Method

Given:

Rate (R) = 5% per annum

Let Principal (P) = ₹100

Since the sum doubles itself, Amount (A) = ₹200

Formula Used:

Simple Interest (SI) = Amount − Principal

SI = (P × R × T) ÷ 100

Calculations:

⇒ SI = 200 − 100 = ₹100

⇒ 100 = (100 × 5 × T) ÷ 100

⇒ 100 = 5 × T

⇒ T = 100 ÷ 5

⇒ T = 20 years

∴ The correct answer is 20 years.

Additional Information

General Formula for n-times

If a sum becomes 'n' times of itself in T years at R% SI, then R × T = (n − 1) × 100.

Simple Interest vs Compound Interest

In Simple Interest, the principal remains constant, whereas in Compound Interest, interest is calculated on the accumulated amount of previous years.

Rule of 72 (Approximate for CI)

For Compound Interest, the time to double a sum is approximately 72 ÷ Rate.

Given:

The average of a, b, and c is 35

The average of a, c, and d is 40

d = 25

Formula used:

Average = (Sum of terms) / (Number of terms)

Calculation:

Average of a, b, and c = 35

⇒ (a + b + c) / 3 = 35

⇒ a + b + c = 35 × 3

⇒ a + b + c = 105

Average of a, c, and d = 40

⇒ (a + c + d) / 3 = 40

⇒ a + c + 25 = 40 × 3

⇒ a + c + 25 = 120

⇒ a + c = 120 - 25

⇒ a + c = 95

Now, we have:

a + b + c = 105

a + c = 95

Subtracting these equations:

(a + b + c) - (a + c) = 105 - 95

⇒ b = 10

∴ The correct answer is option 2.

Given:

Amount after 2 years (A₁) = ₹4,620

Amount after 2.5 years (A₂) = ₹4,725

Time difference = 0.5 years

Formula used:

Simple Interest (SI) = Principal x Rate x Time / 100

SI for time difference = A₂ - A₁

Calculations:

\(SI = A_2 - A_1 = 4725 - 4620 = 105\)

SI for 0.5 years = ₹105

SI for 1 year = \(\frac{105}{0.5} = 210\)

Principal = A₁ - SI for 2 years

\(P = 4620 - 210 \times 2 = 4620 - 420 = 4200\)

Annual rate of interest:

\(SI = \frac{P \times R \times T}{100}\)

\(210 = \frac{4200 \times R \times 1}{100}\)

⇒ \(R = \frac{210 \times 100}{4200} = 5\)

∴ The annual percentage rate is 5%.

Given:

Average weight of a class of 15 students = 30 kg

A new student joins and the average weight increases by 1 kg

Formula Used:

Total weight of n students = Average weight × Number of students

Calculation:

Total weight of 15 students = 30 × 15

Total weight of 15 students = 450 kg

New average weight = 30 + 1 = 31 kg

Total weight of 16 students = 31 × 16

Total weight of 16 students = 496 kg

Weight of the new student = 496 - 450

Weight of the new student = 46 kg

The weight of the new student is 46 kg.

Given:

Numbers are 12, 36, and 80.

Formula Used:

\( \text{LCM}(a, b) = \frac{a \times b}{\text{GCD}(a, b)} \)

For more than two numbers, LCM is calculated iteratively.

Calculation:

First, find the LCM of 12 and 36:

Prime factorization of 12: 2² × 3

Prime factorization of 36: 2² × 3²

LCM of 12 and 36:

⇒ LCM = 2² × 3² = 36

Next, find the LCM of 36 and 80:

Prime factorization of 80: 2⁴ × 5

LCM of 36 and 80:

⇒ LCM = 2⁴ × 3² × 5 = 720

The least common multiple of 12, 36, and 80 is 720.

Given:

Total population = 7,000

Percentage increase in male population = 11%

Percentage increase in female population = 8%

New population = 7,680

Formula used:

New population = Increased male population + Increased female population

Calculations:

Let the number of males = M and females = F

M + F = 7,000

New male population = M × 1.11

New female population = F × 1.08

New population equation:

1.11M + 1.08F = 7680

Substitute M = 7000 - F:

1.11(7000 - F) + 1.08F = 7680

⇒ 7770 - 1.11F + 1.08F = 7680

⇒ 7770 - 7680 = 0.03F

⇒ 90 = 0.03F

⇒ F = \(\frac{90}{0.03}\) = 3000

∴ The number of female citizens in the town is 3,000.

Given:

A and B together can complete a work in 24 days.

B alone can complete the work in 48 days.

Formula Used:

Work done by A and B together in 1 day = 1/24

Work done by B in 1 day = 1/48

Work done by A in 1 day = Work done by A and B together in 1 day - Work done by B in 1 day

Calculation:

Work done by A and B together in 1 day = 1/24

Work done by B in 1 day = 1/48

⇒ Work done by A in 1 day = (1/24) - (1/48)

⇒ Work done by A in 1 day = 2/48 - 1/48

⇒ Work done by A in 1 day = 1/48

Time taken by A to complete the work alone = 48 days

The correct answer is option 1.

Given:

60% of X = 40% of Y

Formula Used:

To find the ratio X : Y

Calculation:

60% of X = 40% of Y

⇒ (60/100)X = (40/100)Y

⇒ 60X = 40Y

⇒ X/Y = 40/60

⇒ X/Y = 2/3

The value of X : Y is 2 : 3.

Given:

Selling Price (SP) = ₹24

Profit Percentage = Cost Price (CP)

Formula used:

Profit Percentage = \(\frac{\text{Profit}}{\text{CP}} × 100\)

Profit = SP - CP

Calculations:

Let CP = x

Profit Percentage = CP = x

\(\frac{\text{SP - CP}}{\text{CP}} × 100 = x\)

⇒ \(\frac{24 - x}{x} × 100 = x\)

⇒ \(\frac{24 - x}{x} = \frac{x}{100}\)

⇒ (24 - x) × 100 = x × x

⇒ 2400 - 100x = x²

⇒ x² + 100x - 2400 = 0

After factorization

⇒ x2  - 20x + 120x - 2400 = 0

(x – 20) (x + 120) = 0

x = 20 and -120

Since CP cannot be negative So, CP = 20

∴ The purchase price of the notebook is ₹20.

Given:

Principal (P) = ₹5,19,500

Rate of Interest (R) = \(7\frac{4}{5}%\) = \(\frac{39}{5} = 7.8%\) = 0.078

Time (T) = 2 years

Let each installment = x

Formula used:

Total Instalments = Principal + 1^st Year Interest + 2^nd Year Interest

Compound Interest Formula:

CI = P × R × T

Where:

P = Principal

R = Rate of Interest

T = Time

Calculations:

1^st Year Compound Interest:

CI₁ = 5,19,500 × 0.078 = 40,521

Remaining principal after the 1^st instalment:

P_remaining = 5,19,500 + 40,521 - x = 5,60,021 - x

2^nd Year Compound Interest:

CI₂ = (5,60,021 - x) × 0.078

CI₂ = 43,681.638 - 0.078x

Total Instalment Equation:

2x = 5,19,500 + 40,521 + 43,681.638 - 0.078x

2x + 0.078x = 6,03,702.638

2.078x = 6,03,702.638

\(x = \frac{6,03,702.638}{2.078} = 2,90,521\)

∴ The equal annual installment is ₹2,90,521.

Given:

Numbers: 16.81 and 12.96

Formula Used:

Mean Proportional = √(a × b)

Calculation:

Let a = 16.81 and b = 12.96

Mean Proportional = √(16.81 × 12.96)

⇒ Mean Proportional = √(217.4976)

⇒ Mean Proportional = 14.76

The mean proportional of 16.81 and 12.96 is 14.76.

Given:

Distance between two places = 88 km

Time to meet (first case) = 8 hours

Akash's speed = a

Vikas's speed = v

In second case:

Akash's speed = \(\frac{7}{6}a\)

Vikas's speed = \(3v\)

Time to meet (second case) = 4 hours

Formula used:

Total Distance = Relative Speed × Time

Calculations:

In the first case:

\((a + v) \times 8 = 88\)

⇒ \((a + v) = \frac{88}{8} = 11\)

In the second case:

\((\frac{7}{6}a + 3v) \times 4 = 88\)

⇒ \((\frac{7}{6}a + 3v) = \frac{88}{4} = 22\)

Substituting \(a + v = 11\) into the second equation:

⇒ \(\frac{7}{6}a + 3(11 - a) = 22\)

⇒ \(\frac{7}{6}a + 33 - 3a = 22\)

⇒ \(-\frac{11}{6}a + 33 = 22\)

⇒ \(-\frac{11}{6}a = 22 - 33 = -11\)

⇒ \(a = \frac{6 \times 11}{11} = 6\)

Substitute \(a = 6\) into \(a + v = 11\):

⇒ \(6 + v = 11\)

⇒ \(v = 5\)

∴ Akash's speed = 6 km/h, Vikas's speed = 5 km/h.